Metamath Proof Explorer


Theorem spcegf

Description: Existential specialization, using implicit substitution. (Contributed by NM, 2-Feb-1997)

Ref Expression
Hypotheses spcgf.1 ⊢ Ⅎ 𝑥 𝐴
spcgf.2 ⊢ Ⅎ 𝑥 𝜓
spcgf.3 ⊢ ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) )
Assertion spcegf ( 𝐴 ∈ 𝑉 → ( 𝜓 → ∃ 𝑥 𝜑 ) )

Proof

Step Hyp Ref Expression
1 spcgf.1 ⊢ Ⅎ 𝑥 𝐴
2 spcgf.2 ⊢ Ⅎ 𝑥 𝜓
3 spcgf.3 ⊢ ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) )
4 2 nfn ⊢ Ⅎ 𝑥 ¬ 𝜓
5 3 notbid ⊢ ( 𝑥 = 𝐴 → ( ¬ 𝜑 ↔ ¬ 𝜓 ) )
6 1 4 5 spcgf ⊢ ( 𝐴 ∈ 𝑉 → ( ∀ 𝑥 ¬ 𝜑 → ¬ 𝜓 ) )
7 6 con2d ⊢ ( 𝐴 ∈ 𝑉 → ( 𝜓 → ¬ ∀ 𝑥 ¬ 𝜑 ) )
8 df-ex ⊢ ( ∃ 𝑥 𝜑 ↔ ¬ ∀ 𝑥 ¬ 𝜑 )
9 7 8 imbitrrdi ⊢ ( 𝐴 ∈ 𝑉 → ( 𝜓 → ∃ 𝑥 𝜑 ) )